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    Forces Acting on an Object

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    Three forces acting on an object are given by F1 = (−1.85 i + 7.35 j) N, F2 = (4.70 i − 1.15 j) N, and F3 = (−45.5 i) N. The object experiences an acceleration of magnitude 3.70 m/s2.
    (a) What is the direction of the acceleration?
    (degrees counterclockwise from the positive x axis)
    (b) What is the mass of the object in kg?
    (c) If the object is initially at rest, what is its speed after 16.0 s in m/s?
    (d) What are the velocity components of the object after 16.0 s in m/s?

    © BrainMass Inc. brainmass.com December 24, 2021, 11:12 pm ad1c9bdddf
    https://brainmass.com/physics/newtons-laws/forces-acting-object-547720

    SOLUTION This solution is FREE courtesy of BrainMass!

    F1 = (-1.85 i + 7.35 j) N,
    F2 = (4.70 i - 1.15 j) N,
    F3 = (-45.5 i) N.

    Net force,
    Fnet = F1 + F2 + F3
    = (-1.85 i + 7.35 j) + (4.70 i - 1.15 j) + (-45.5 i)
    = (-1.85 + 4.70 - 45.5)i + (7.35 - 1.15)j
    = (-42.65 i + 6.20 j) N == (Fx i + Fy j)

    (a) Direction of force and acceleration will be the same = ?
    Let us assume, angle with +x axis = theta (counter clock wise)
    tan(theta) = Fy/Fx = 6.20/(-42.65) = -0.145
    => theta = arctan(-0.145) = 180 - 8.27 = 171.73 degree

    (b) mass (m) = ?
    |F| = m*|a|
    |a| = 3.7 m/s^2
    |F| = sqrt(42.65^2 + 6.2^2) = 43.1 N

    Hence,
    m = |F|/|a| = 43.1/3.7 = 11.6 kg

    (c)
    Initial velocity, u = 0
    acceleration, a = 3.7 m/s^2
    time, t = 16.0 s
    velocity, v = ?
    v = u +at
    v = 0 + 3.7*16 = 59.2 m/s

    (d)
    x-component of velocity,
    v_x = v*cos(theta) = 59.2*cos(171.73) = -58.58 m/s

    y component of velocity,
    v_y = c*sin(theta) = 59.2*sin(171.73) = 8.52 m/s

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:12 pm ad1c9bdddf>
    https://brainmass.com/physics/newtons-laws/forces-acting-object-547720

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